 we had in the last lecture we had tried to find out the inductance the mutual inductance between the rotor coil and the stator coil for a cylindrical rotor salient pole cylindrical stator salient pole rotor arrangement and in order to see how that expression for that can be derived we had a look at how the flux distribution inside this kind of machine would look like. We have then understood that the flux distribution inside the machine for the rotor angle equal to 0 the rotor angle equal to 0 was the case where this rotor is positioned horizontally you have a coil on the stator here one single coil on the stator the rotor position horizontally and you have coils found on the rotor and for that location of the rotor if you travel around the air gap around this inner circumference of the stator and try to find out how the flux density variation would look like we saw that the flux density variation looks like this the flux density is almost 0 for some time when the pole phase arrives you have a large value of flux density and once you cross the pole phase the flux density is again 0 and the next pole phase the flux density is negative and so on that is the kind of waveform that you have if the rotor is going to move we find that this flux density waveform has bodily shifted along the x axis but the wave shape remains the same the first waveform is drawn for an angle ? r equal to 0 the second waveform is drawn for an angle ? r equal to 30 degrees the waveform has moved by 30 degrees at 60 degrees we find that the waveform has moved further compared to this and the wave shape essentially remains the same except for the small dip and when the rotor has moved by 90 degrees we find that the position of the dip is same but the waveform has moved by another 30 degrees so this essentially how the flux density around the air gap this x axis denotes the flux density variation as you travel around the air gap of the machine this dip that occurs is due to the slot which is present here you see that there is a slot present if you start from the circumference angle 0 here and then you travel all the way here so at an angle equal to 180 degrees you find that there is a slot and because of that slot the flux density therefore dips and when the waveform moves further the position of the dip remains the same this in effect is due to the slotting and we said that we are going to neglect slotting in this evaluation and therefore the flux density waveform is approximated as a square wave like this with an intervening 0 intervals now this waveform is again drawn for the case of rotor angle equal to 0 degrees as the rotor angle moves this waveform is going to shift along the x axis as we had seen earlier so now the issue is having known that the flux density waveform is going to look like this quasi square wave with 0 intervening intervals how do we go about deriving an expression for the mutual inductance now we note here that the flux density waveform that we have drawn is the flux density that arises due to excitation in the rotor so the rotor is being excited and for a given position in the rotor one can determine what the flux density waveform is as you travel around the circumference of the air gap we want to find the mutual inductance therefore the mutual inductance m between stator and rotor is defined as the flux linkage in the stator coil per unit current flowing in the rotor so that is what we need to find out so now the question is flux density around the air gap is a function of which is to say is dependent on the circumferential angle by circumferential angle what we mean is the angle that you would traverse if you travel along the inner circumference of the stator that is your circumferential angle we start with 0 here and then travel around reach 180 here and then back to 360 here that is a circumferential angle and we have seen that it is dependent on the circumferential angle for some place it is 0 then maximum flux then 0 and so on right so if this is a case how do we find out what is the flux linkage of this coil on the stator so to find out flux linkage flux linkage is defined as the number of turns in the stator coil multiplied by the flux passing through passing through an area spanned by the coil now the area spanned by the coil what we can conveniently take we know that flux lines are going from the rotor and going to cross into the stator and therefore the area spanned by the coil could be taken as the area along the inner circumference of the stator from one side of the coil to the other side of the coil this entire inner circumferential area is what we can consider what we mean by the circumferential area is that this machine is extending along the axis of itself which lies perpendicular to the plane of the drawing and therefore it is basically a cylindrical structure which rises from which goes out of this plane and if it is going to be a cylinder then there is a circumferential area of the cylinder inside flux is passing from the rotor into the stator along the circumferential area and therefore that is a circumferential area which we are going to consider in order to determine the flux that is being linked by this loop and so how does one do that we know that flux is given by the flux density multiplied by da that is flux density into area is going to give you flux that is passing through that area so if you now consider an elemental area let us say that we are looking at a small area here from the center let us say that this area this arc subtends an angle d a and which lies at some angle a from your reference 0 degrees so if you want to find out the flux that is passing through this elemental area that is obtained by the expression b at this value of a b as a function of a multiplied by d a that is at this point and this will then if you multiply this elemental flux linkage that is the flux linkage arising due to the flux passing through this elemental area we can call it as d ? s this d ? s is then equal to the number of turns in the stator multiplied by b.da okay. Now in order to work out this example let us consider a small modification which will help us to understand winding better in the drawing that is shown here you have one coil side that is placed here and the other coil side lies exactly 180 degrees away now in some cases this may not be the situation the other coil side may not lie 180 degrees away but it may lie in a slot that is probably less than 180 degrees away may be a slot lying here you would have this other coil side here in which case we call this angle this angle let us call this as ? we call ? as the coil span in other words the coil spans an angle of ? so let us say that this coil that we are considering is now spanning an angle of ? which is likely to be less than 180 degrees and we want to find out the flux linkage in this loop. So as we said we are now trying to find out how much flux is linked by this coil by finding out how much flux is going to cross the circumferential surface of the stator inner circumferential surface this surface area how much flux is going to cross so in order to do that we have started considering an elemental area we have found out the flux linkage now this expression this approach can be used but we first have to have an expression for B and B as you know we have approximated the wave form of B to look like this it is a quasi square wave which is going here and we have this coil which is now situated one side of the coil is here another side of the coil is somewhere here which is less than 180 degrees okay. So in order to do this what we can do since we need to have a functional form describing the variation of B with respect to a we can try to resolve the wave form of B the spatial distribution of B into Fourier series into its Fourier series expansion then we will have B of a is equal to some sigma of let us say a n cos n a plus bn sin n a sigma going from n equal to 1 to 8 and then one can use this expression to substitute in this value of B and derive an expression for d ?s so how does one go about doing this so we have seen that this wave form of B is this wave form is going to be an odd function of this angle in the sense if you now extend this wave form to the left of 0 degrees then this wave form is going to look like this which means it is an odd function of the circumferential angle and therefore if we try to derive the Fourier series expression for this expression will contain only the sinusoidal terms the cos sinusoidal terms which are a n terms will not be there therefore all a n are equal to 0 in this wave form we need to determine only the terms corresponding to bn and further we note that this wave form has half wave symmetry in the sense the wave form becomes negative after every half a cycle and therefore we have the condition f of a is equal to – of f of a plus ? that is your half wave symmetric wave form and therefore if this is the case we also know that only odd harmonics exist right and therefore we can now write an expression for bn is equal to 2 by 2 ? integral 0 to 2 ? b of a sin n a d a and since this wave form is half wave symmetric one can simplify this as 4 by 2 ? into integral 0 to ? b of a sin n a d a and further this wave form also has a symmetry about 90 degrees you find that if you draw the angle draw the line corresponding to 90 degrees the wave form on either side are equal this forms a line of symmetry and therefore you can further simplify this expression as 8 by 2 ? integral 0 to ? by 2 b of a sin n a d a now in doing the integration from 0 to ? by 2 we further note that this wave form is 0 over some angle let us call this angle for which it is 0 from 0 degrees let us call this angle as ? then what you have is this integral reduces to 8 by 2 ? integral from ? to ? by 2 b of a sin n a d a and in the range ? to ? by 2 b has a certain maximum value which is constant and therefore we can write it as b hat divided by 2 ? integral ? to ? by 2 sin n a d a and so let us simplify this expression this is nothing but 8 b hat by 2 ? integral of sin n a is cos n a by n limits going from ? to ? by 2 and this is therefore 8 b hat by 2 n ? cos n ? by 2 – cos n ? by n ? that is your expression that you have now note that we have already observed that this expression in this expression you will have only odd values of n because this has only odd ordered harmonics even ordered harmonics do not exist in the wave form so it is enough if you consider this for odd values of n and if you put odd values of n here let us say n equal to 1 then this expression that is b1 is then 8 b hat by 2 ? into cos of ? by 2 is 0 so you have – cos ? and then for n equal to 3 you have b3 equals 8 b hat by 6 ? into cos of 3 ? by 2 is again 0 so you have – cos 3 ? and then you have n equal to 5 you have b5 equals 8 b hat by 10 ? cos of 5 ? by 2 is cos of 5 ? by 2 is equal to cos of 2 ? plus ? by 2 which is equal to cos ? by 2 and therefore that is 0 so you have – cos 5 ? and therefore 1 can see that b of a can then easily be written as 8 b hat by 2 ? – cos ? into sin a plus 8 b hat by 6 ? into – cos 3 ? into sin 3 a plus 8 b hat by 10 ? into – cos 5 ? into sin 5 a plus so on this is the series that you get now one can simplify this put it in a more simpler notation for b hat by ? into ? n equals 1 to 8 we will take the – sign also outside then you have cos of 2n – 1 into ? into sin of 2n – 1 into a divided by 2n – 1 so for n equal to 1 you have 2 into 1 – 1 that is cos ? for n equal to 2 you have cos 3 ? for n equal to 3 you have cos 5 ? and the denominator also appropriately vary. So this is the expression for b of a note that in this expression n goes from 1 to 8 we have taken care of the odd harmonics by putting 2n – 1 here so this is then your expression for b a and what we want to do is to substitute this expression for b a in this expression for d ? and therefore what we get is d ? s is nothing but – 4 b hat by ? multiplied by ns n equal to 1 to 8 cos of 2n – 1 ? sin 2n – 1 a by 2n – 1 this is n into b a you have to multiply this by d a. Now in this machine geometry that you have what would happen with the flux lines is that these flux lines would go from the stator from the rotor to the stator we are exciting the stator and trying to find out what is the flux linkage in the rotor at this interface we would find that the flux lines enter the stator in such a manner that flux lines are perpendicular to stator surface stator inner surface which means that the flux density vector b is normal to the area at the inner circumference flux lines are perpendicular to the stator inner surface and therefore the flux density vector is normal to the area at the inner circumference and therefore b ? d a will simply reduce to b multiplied by d a and what is this d a d a is the elemental area around the inner circumference of the stator which means it is basically the arc length here multiplied by the length along the axis and therefore if we say that the length along the axis is L the total area of that elemental segment is L into r d a this is nothing but your d a that is there. So if you substitute this expression for d a then what you have is L r d a so this is now an expression for the elemental flux linkage due to flux passing over this elemental area only through this area but actually the area through which flux is going to pass which links this coil is this entire area and therefore in order to find out what is the flux linkage due to all the flux passing through this area we then need to integrate so the flux linkage ? s is integral of d ? s and the limits of the integral are from 0 a equal to 0 to a equal to the other end of this coil of the stator note that the other end of the coil lies at an angle ? and therefore you integrate over a going from 0 to ? and so this is a going from 0 to ? so let us do this integration now in this obviously this 4 b ns they are all constant so you have – 4 b hat ns multiplied by L x r divided by ? x integral a going from 0 to ? cos I forgot the summation so let us put the summation also there ? n equal to 1 to ? you have cos of 2n – 1 ? by 2n – 1 and integral a equal to 0 to ? sin of 2n – 1 a sorry that is a d a so that is the integration that we need to do so that can be written as – 4 b hat ns L x r by ? x ? n equal to 1 to ? cos 2n – 1 into ? by 2n – 1 integral of sin is again – cos so you get a – sin here and then you have cos of 2n – 1 into ? by 2n – 1 this going from 0 to ? and therefore this expression is – 4 b hat ns x L x r divided by ? n equal to 1 to ? – cos of 2n – 1 ? by 2n – 1 whole square and now you have cos of 2n – 1 ? – cos of 2n – 1 into 0 and 2n – 1 into 0 is 1 so you have basically – 4 b hat ns L r by ? n equal to 1 to ? – cos of 2n – 1 ? by 2n – 1 whole square cos of 2n – 1 into ? so this is the expression that we land up with for the flux linkage now note that this expression is however valid only for one particular position of the rotor that is the rotor having ? r equal to 0 if ? r equal is not equal to 0 what do we do how do we describe the flux density waveform for the case where ? r is not equal to 0 now if ? r is not equal to 0 what we have seen earlier is that the flux density waveform is going to bodily shift along the x axis so you have ? r equal to 0 and then ? r equal to 30, 60, 90 the waveform is shifting so essentially therefore if you want to have the flux density waveform at which is going to vary as a function of a but not only as a function of a it is also going to vary as a function of the rotor angle this is nothing but the waveform shifting as the rotor angle moves and this is therefore the same as b a – ? r you say that if you have a waveform f of x and you want to shift it by some unit then what you have is f of x – ? that is the same thing we are doing here and this is a – ? r therefore if you want to describe it like this what you have to do is in these expressions instead of a you will have to put a – ? r and therefore now the flux density expressions would become b of a, ? r will be – 4 b hat ns x l x r divided by ? multiplied by ? going from n equal to 1 to 8 – cos 2n – 1 ? 2n – 1 ? x sin 2n – 1 x now a – ? r divided by 2n – 1 this would then be your flux density distribution if you include ? r also this is therefore a 2 variable expression now this would then give us the variation of flux linkage as the rotor angle also is going to change right this expression earlier we had derived a simpler form of the expression so that we get to grips with the integration process that is involved and here we had fixed the rotor at a particular position the particular position was ? r equal to 0 so if ? r equal to 0 one obviously sees that there is no ? r here the expression becomes the same as earlier case right so what we really have to do is now to get do the same integration with this expression and therefore what you have is integral 0 to ? d ? s is your expression for ? s and this is then equal to – 4 b hat ns x l x r divided by ? x s n equal to 1 to 8 cos 2n – 1 x ? by 2n – 1 x integral 0 to ? of sin 2n – 1 x ? r multiplied by d ? so that is the integration that we would have to do if you want to consider the rotor angle variation also and this integration therefore is – 4 b hat ns x l x r divided by ? s n equal to 1 to 8 cos 2n – 1 x ? by 2n – 1 x from sin you would get a – cos so you get a – and then this is cos of 2n – 1 x a – ? r divided by 2n – 1 so we put the square here and this part goes from 0 to ? so let us simplify the notation a little we will call as k 2n – 1 as 4 times ? hat ns x l x r divided by ? x cos of 2n – 1 x cos of 2n – 1 ? divided by 2n – 1 whole square so that it is easier for us to deal with so this is then ? of n equal to 1 to 8 k 2n – 1 x cos of 2n – 1 x ? – ? r – cos of 2n – 1 x 0 – ? r that is – ? r which therefore reduces to ? n equal to 1 to 8 k 2n – 1 x cos 2n – 1 x ? – ? r – cos of 2n – 1 ? so let us expand the first term and that would give us cos of 2n – 1 x ? x cos of 2n – 1 x ? r – sin of 2n – 1 x ? x sin of 2n – 1 x ? r this is an expansion for the first 2 terms and then you have – cos of 2n – 1 x ? r so one can group the first and last terms together this is nothing but cos of 2n – 1 x ? r x cos 2n – 1 x ? – 1 x sin of 2n – 1 x ? sin 2n – 1 x ? r okay so what we essentially see from this expression is that let us maybe it would be of it would be instructive to write down the first few expression of the sequence the first few terms of the sequence would then be for n equal to 1 what we have arrived at is the flux linkage for n equal to 1 k 2n – 1 is nothing but 4 ? hat ns so 4 ? hat ns x l x r by ? this is the constant term and then for n equal to 1 what we have from here is cos ? by 1 square so you have cos ? and then this term would give us cos of ? r into may be good to use the earlier expression itself perhaps cos of ? – ? r – cos of ? r is the first expression and then you would have cos of 3 ? x cos of cos of 3 ? x cos of 3 times ? – ? r – cos 3 ? r and then you will have cos of 5 ? of course this would be divided by 9 this would be divided by 25 cos of 5 ? x cos of 5 times ? – ? r – cos 5 ? r and so on this would be see it now one can see that this expression contains a fundamental term which corresponds to the variation with respect to ? itself and then it contains a third harmonic which means there is a variation with respect to 3 times ? and then there is a fifth harmonic which contains variation with respect to 5 times ? and so on there will be many more harmonics. And however we find that the harmonic amplitudes go down rather fast the third harmonic is one ninth of what would have been the first the fifth harmonic is one 25th of what would have been the first and therefore in the flux linkage expression the harmonic terms are going to contribute less and less as the harmonic order increases and because of this reason it is most in most situations sufficient to look at the fundamental component alone right and there also we have derived this expression for a certain angle ? now let us look at harmonic number 3 that has a variation as cos of 3 times ? – ? r – cos 3 ? r. Now supposing you choose ? equal to 2 ? x 3 then what happens to this term is it becomes cos of 3 times 2 ? x 3 – ? r – cos 3 ? r which is the same as cos of 2 ? – 3 ? r – cos 3 ? r which is the same as cos of 3 ? r – cos 3 ? r which is 0 so it means that you can choose a value of ? in such a way that a specific harmonic in the flux linkage is eliminated. So this is one of the advantages of having ? not equal to 180 degrees okay so appropriately one can choose an angle of ? so having understood that even if you do not choose an angle of equal to 2 ? x 3 the amplitudes are going to diminish as 1 x n2 so in most cases for our analysis where we encounter non-sinusoidal distributions which are typically going to be the case what we would do is consider only the fundamental of the distribution. Therefore if you consider the fundamental component of the flux linkage that can then be written as 4 b hat x ns x l x r x ? x cos ? x cos of ? – ? r – cos ? r this is in most cases this expression would be sufficient for us to get reasonably good result. Having understood that let us now look at the case where ? is equal to ? if ? is equal to ? it means that the other end of the coil other side of the loop that we are considering lies 180 degree away and these kind of arrangements is this is called as a full pitched coil. So for a full pitched coil then the flux linkage expression would be ? s is equal to 4 ? hat ns x l x r divided by ? cos ? cos of ? – ? r is – cos ? r so this results in this is x – 2 times cos ? r so you end up with – 8 b hat ns x l x r divided by ? cos ? cos ? r which is some maximum value multiplied by ? maximum or ? hat as per our notation x cos ?. So having derived this expression then let us look in the next lecture session how to deal with this and how to get the mutual inductance out of it should be fairly easy to do that b hat is a function of is and therefore if you divide this by is you get an expression for the mutual inductance we will look at that in the next lecture session.